Past exam of the differential equations 1 course of the University of Oxford / 2017 / 3 / b / Solution

In $x,y>0$, the characteristic equation of a partial differential equation gives $dx/\sqrt{x}=\pm \sqrt{y}dy$. Integration yields the characteristic coordinatesTheir Jacobian determinant is $-\sqrt{y}/(2\sqrt{x})\ne 0$, so they are valid independent coordinates. To see cancellation of the lower-order terms explicitly, first put $r=\sqrt{x}$, $s=y^{3/2}/3$ and $u(x,y)=U(r,s)$. The chain rule givesSubstitution cancels the first-derivative terms, leaving $U_{ss}-U_{rr}=0$. With $\xi =r+s$, $\eta =r-s$ and $U=V(\xi ,\eta )$, the characteristic normal form is $V_{\xi \eta }=0$. Integrating once in each variable giveswhere $g,h$ are arbitrary twice continuously differentiable functions on the relevant coordinate intervals. This is a curvilinear reduction to the one-dimensional wave equation.